Does the Laplacian commutes with elements of the basis of the Lie algebra?

Question

Let $G$ be a compact Lie group with Lie algebra $\mathfrak{g}$. I know that if $g$ is semi-simple then the Laplace-Beltrami operator on $G$ agrees with the Casimir element and therefore commutes with with the elements of $\mathfrak{g}$. Here we use the Riemmanian metric given by the Killing form.

In the general case, we can write $\mathfrak{g} = \mathfrak g' \oplus \mathfrak g''$ with $\mathfrak g'$ semi-simple. Now we can endow $G$ with a Riemmaniam metric given by the Killing form on $\mathfrak g'$ and any other metric on $\mathfrak g''$.

Is it possible to show that in this case the Laplace-Beltrami operator still commutes with the elements of $\mathfrak g$?

If this is a well known result, what is a good reference for this?

Answer 1

In general, no.

Let $E$ denote the restriction of the Laplace-Beltrami operator to the identity, and look at it's action on functions. Then if coincides with the Laplace-deRham operator, $\delta d$. If we have basis, $E_i$, then $E=E^iE_i$, and let $[E, E_j]=a_i^kE_k$ and $[E, E^j]=b_k^jE^k$. Then $[E_i, E_jE^j]=[E_i, E_j]E^j+E_j[E_i, E^j]=a_i^kE_kE^j+b_k^jE_jE^k$. So that we are reduced to the relation, $a_i^k=-b_k^j$. Now we compute, denoting the metric as $B$, $B([E_j, E], E^l)-B(E_j, [E, E^l])=-a_j^l-b_l^j$. Using linearity we thus see that community of $E$ with $\mathfrak{g}$ is equivalent to the relation $B([X, Y], Z)=B(X, [Y, Z])$.

Every such form is a scalar multiple of the Killing form when $\mathfrak{g}$ is simple, so it is determined there. When $\mathfrak{g}$ is abelian, it is likewise possible since the relation is redundant. That said, there are many examples of Lie algebras that do not possess such a form, being known as quadratic lie algebras. For example take two vectors, $X, Y$, with $[X, Y]=X$. Then $B(X, Y)=B([X, Y], Y)=B(X, 0)=0$ and $B(X, X)=B(X, [X, Y])=B([X, X], Y)=0$ so that this cannot be non-degenerate.